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NNADVOKAT [17]
3 years ago
8

Pls Helpp meee Which one is it??? I will give u brainliest

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

Answer:

The answer will be the last one which means that will 2

Step-by-step explanation:

You need to find

(f.g)=2 or (f.g) (-2)=2

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The odometer in Mr. Washington's car does not work correctly. The odometer recorded 13.2 miles for
IrinaVladis [17]

I think  the answer its 1.8 percent because if you subtract 15-13.2 comes out to be 1.8.

but l also think it could be 12 percent error.      

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2 years ago
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AfilCa [17]

Answer:

The answer would be D. 5

Step-by-step explanation:

4 0
2 years ago
2. In how many ways can 3 different novels, 2 different mathematics books and 5 different chemistry books be arranged on a books
insens350 [35]

The number of ways of the books can be arranged are illustrations of permutations.

  • When the books are arranged in any order, the number of arrangements is 3628800
  • When the mathematics book must not be together, the number of arrangements is 2903040
  • When the novels must be together, and the chemistry books must be together, the number of arrangements is 17280
  • When the mathematics books must be together, and the novels must not be together, the number of arrangements is 302400

The given parameters are:

\mathbf{Novels = 3}

\mathbf{Mathematics = 2}

\mathbf{Chemistry = 5}

<u />

<u>(a) The books in any order</u>

First, we calculate the total number of books

\mathbf{n = Novels + Mathematics + Chemistry}

\mathbf{n = 3 + 2 +  5}

\mathbf{n = 10}

The number of arrangement is n!:

So, we have:

\mathbf{n! = 10!}

\mathbf{n! = 3628800}

<u>(b) The mathematics book, not together</u>

There are 2 mathematics books.

If the mathematics books, must be together

The number of arrangements is:

\mathbf{Maths\ together = 2 \times 9!}

Using the complement rule, we have:

\mathbf{Maths\ not\ together = Total - Maths\ together}

This gives

\mathbf{Maths\ not\ together = 3628800 - 2 \times 9!}

\mathbf{Maths\ not\ together = 2903040}

<u>(c) The novels must be together and the chemistry books, together</u>

We have:

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the novels in:

\mathbf{Novels = 3!\ ways}

Next, arrange the chemistry books in:

\mathbf{Chemistry = 5!\ ways}

Now, the 5 chemistry books will be taken as 1; the novels will also be taken as 1.

Literally, the number of books now is:

\mathbf{n =Mathematics + 1 + 1}

\mathbf{n =2 + 1 + 1}

\mathbf{n =4}

So, the number of arrangements is:

\mathbf{Arrangements = n! \times 3! \times 5!}

\mathbf{Arrangements = 4! \times 3! \times 5!}

\mathbf{Arrangements = 17280}

<u>(d) The mathematics must be together and the chemistry books, not together</u>

We have:

\mathbf{Mathematics = 2}

\mathbf{Novels = 3}

\mathbf{Chemistry = 5}

First, arrange the mathematics in:

\mathbf{Mathematics = 2!}

Literally, the number of chemistry and mathematics now is:

\mathbf{n =Chemistry + 1}

\mathbf{n =5 + 1}

\mathbf{n =6}

So, the number of arrangements of these books is:

\mathbf{Arrangements = n! \times 2!}

\mathbf{Arrangements = 6! \times 2!}

Now, there are 7 spaces between the chemistry and mathematics books.

For the 3 novels not to be together, the number of arrangement is:

\mathbf{Arrangements = ^7P_3}

So, the total arrangement is:

\mathbf{Total = 6! \times 2!\times ^7P_3}

\mathbf{Total = 6! \times 2!\times 210}

\mathbf{Total = 302400}

Read more about permutations at:

brainly.com/question/1216161

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nadya68 [22]
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Done&lt;<br> In the diagram below, AB BC, and mZBCD = 115°. Find mZB.<br> Statement<br> B.
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B.

I took the test :)
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3 years ago
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